Answer
$t^{\circ}_{h}=t^{\circ}-0.01^{\circ}/m\cdot h$ for $0\leq h \leq 12 000\text{ m}.$
Work Step by Step
Air temperature drops about $1^{\circ}$ for each $100$-$m$ or $0.01^{\circ}$ for each meter. So if the ground temperature $t^{\circ}$ and you climbed $h$ meters, the temperature at a height of $h$ meters will be
$t^{\circ}-0.01^{\circ}/m\cdot h$, where $h\geq0$ and $h\leq 12 000$ m or $12$ km.
